AP Chemistry Unit 5: Kinetics Practice Exam

Assessment for Unit 5: Kinetics

Estimated Time: 60 minutes

Exam Details

Questions: 26 total (25 multiple choice, 1 short answer)
Time: 60 minutes (recommended)
Topics Covered: Unit 5: Kinetics
Format: Multiple Choice, Short Answer questions matching real exam difficulty
Scoring: Instant results with detailed explanations

A. Multiple-Choice Questions (25 Questions)

Select the one best answer for each question.

1. [Skill: 2.A | Topic: 5.1] A student studies the reaction $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$ The concentration of $NO_2$ is measured over the first 60.0 s of the reaction. Time (s): 0.0, 60.0 $[NO_2]$ (M): 0.100, 0.070 Based on the data, what is the average rate of formation of $O_2$ over the first 60.0 s?

(A)$5.0\times10^{-4}\ \text{M s}^{-1}$(B)$2.5\times10^{-4}\ \text{M s}^{-1}$(C)$1.5\times10^{-4}\ \text{M s}^{-1}$(D)$7.5\times10^{-4}\ \text{M s}^{-1}$

Refer to the figure below.

2. [Skill: 4.A | Topic: 5.1] A student investigates how surface area affects reaction rate using the reaction $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$. Two trials use the same mass of Mg and the same $[HCl]$ at the same temperature. Trial 1: Mg ribbon (one piece) Trial 2: Mg powder (many small particles) The student plots the volume of $H_2(g)$ collected versus time for both trials. Which curve (1 or 2) best represents Trial 2 (Mg powder)?

(A)Curve 1, because Mg powder produces fewer collisions due to smaller particles.(B)Curve 1, because Mg powder has lower concentration than Mg ribbon.(C)Curve 2, because Mg powder has a larger surface area and therefore a greater initial rate (steeper initial slope).(D)Curve 2, because Mg powder increases the reaction temperature, increasing the rate.

3. [Skill: 3.B | Topic: 5.1] A reaction is conducted in three separate trials using the same initial concentrations of reactants. Trial 1: 25°C, no catalyst Trial 2: 35°C, no catalyst Trial 3: 25°C, catalyst added Which statement best explains why Trial 2 and Trial 3 each have a greater initial rate than Trial 1?

(A)Trial 2 has a higher initial rate because it has more reactant particles per unit volume; Trial 3 has a higher initial rate because catalysts increase the concentration of reactants.(B)Trial 2 has a higher initial rate because a higher temperature increases the fraction of collisions with sufficient energy; Trial 3 has a higher initial rate because a catalyst provides an alternative pathway with lower activation energy.(C)Trial 2 has a higher initial rate because higher temperature changes the stoichiometry of the balanced equation; Trial 3 has a higher initial rate because a catalyst increases the enthalpy change of the reaction.(D)Trial 2 has a higher initial rate because higher temperature increases activation energy; Trial 3 has a higher initial rate because a catalyst shifts equilibrium toward products.

Refer to the figure below.

4. [Skill: 4.A | Topic: 5.10] A reaction proceeds by a two-step mechanism. The reaction energy profile is shown. [Image Cue]: Reaction energy profile diagram, Title: "Two-step reaction energy profile". y-axis: Potential energy (kJ/mol). x-axis: Reaction coordinate. Reactants at 50 kJ/mol. First transition state peak at 90 kJ/mol. Intermediate valley at 20 kJ/mol. Second transition state peak at 60 kJ/mol. Products at 10 kJ/mol. Which of the following gives the correct activation energy for step 1, activation energy for step 2, and overall enthalpy change, respectively?

(A)$E_{a,1}=40\ \text{kJ/mol},\ E_{a,2}=40\ \text{kJ/mol},\ \Delta H=-40\ \text{kJ/mol}$(B)$E_{a,1}=40\ \text{kJ/mol},\ E_{a,2}=10\ \text{kJ/mol},\ \Delta H=-40\ \text{kJ/mol}$(C)$E_{a,1}=90\ \text{kJ/mol},\ E_{a,2}=60\ \text{kJ/mol},\ \Delta H=-10\ \text{kJ/mol}$(D)$E_{a,1}=70\ \text{kJ/mol},\ E_{a,2}=50\ \text{kJ/mol},\ \Delta H=+40\ \text{kJ/mol}$

5. [Skill: 2.A | Topic: 5.10] A proposed two-step mechanism has the following potential energies (relative to reactants) for key points along the reaction coordinate: - Reactants: 0 kJ/mol - Transition state 1: 75 kJ/mol - Intermediate: 30 kJ/mol - Transition state 2: 110 kJ/mol - Products: 20 kJ/mol Which statement correctly gives the activation energy for step 2 and the overall enthalpy change for the reaction?

(A)$E_{a,2}=80\ \text{kJ/mol}$ and $\Delta H=+20\ \text{kJ/mol}$(B)$E_{a,2}=110\ \text{kJ/mol}$ and $\Delta H=+20\ \text{kJ/mol}$(C)$E_{a,2}=80\ \text{kJ/mol}$ and $\Delta H=-20\ \text{kJ/mol}$(D)$E_{a,2}=35\ \text{kJ/mol}$ and $\Delta H=+110\ \text{kJ/mol}$

Refer to the figure below.

6. [Skill: 1.B | Topic: 5.10] A reaction occurs in two elementary steps and has the following energetic features: - The overall reaction is exothermic. - The intermediate is at a higher potential energy than the reactants. - The second step has a larger activation energy than the first step. Which reaction energy profile best matches these features? [Image Cue]: Four small reaction energy profiles labeled (A)–(D). Each has two peaks (two transition states) with one intermediate valley. All have y-axis Potential energy and x-axis Reaction coordinate. - Profile (A): Intermediate below reactants; products below reactants; second peak lower (relative to its preceding valley) than first. - Profile (B): Intermediate above reactants; products below reactants; second peak higher above the intermediate than the first peak is above reactants. - Profile (C): Intermediate above reactants; products above reactants; second activation barrier smaller than first. - Profile (D): Intermediate below reactants; products above reactants; second activation barrier larger than first.

(A)Profile (A)(B)Profile (B)(C)Profile (C)(D)Profile (D)

Refer to the figure below.

7. **1.** [Skill: 5.A | Topic: 5.11] A student investigates a reaction that can occur with or without a catalyst. The potential energy diagram for both pathways is shown. [Image Cue]: Potential energy diagram, "Catalyzed vs. Uncatalyzed Reaction Pathway". y-axis: Potential Energy; x-axis: Reaction Coordinate. Two curves start at the same reactant energy and end at the same product energy. The uncatalyzed curve has one high peak. The catalyzed curve has a lower maximum peak (may be shown as one lower peak or multiple smaller peaks). The overall energy difference between reactants and products is the same for both pathways. Which of the following best describes the effect of adding the catalyst?

(A)The catalyst lowers the activation energy for both the forward and reverse reactions but does not change the overall energy difference between reactants and products.(B)The catalyst lowers the activation energy for the forward reaction only and increases the activation energy for the reverse reaction.(C)The catalyst increases the potential energy of the reactants, making the reaction more exothermic and faster.(D)The catalyst increases the energy difference between reactants and products, which increases the rate constant.

8. **2.** [Skill: 1.B | Topic: 5.11] The decomposition of hydrogen peroxide can be catalyzed by iodide ion. A proposed mechanism is shown. Step 1 (slow): $H_2O_2 + I^- \rightarrow H_2O + IO^-$ Step 2 (fast): $H_2O_2 + IO^- \rightarrow H_2O + O_2 + I^-$ Which statement correctly identifies the catalyst and a reaction intermediate in the mechanism?

(A)Catalyst: $IO^-$ ; Intermediate: $I^-$(B)Catalyst: $I^-$ ; Intermediate: $IO^-$(C)Catalyst: $H_2O_2$ ; Intermediate: $IO^-$(D)Catalyst: $I^-$ ; Intermediate: $O_2$

9. **3.** [Skill: 6.A | Topic: 5.11] A student studies the hydrogenation of ethene on a platinum catalyst: $C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ Two trials are performed at the same temperature, with the same initial amounts of gases. - Trial 1 uses a single piece of platinum foil. - Trial 2 uses the same mass of platinum in a finely divided powder. The student observes that Trial 2 proceeds significantly faster than Trial 1. Which of the following best explains the observation?

(A)The powdered platinum provides greater surface area, increasing the number of adsorption sites and enabling additional elementary steps involving surface-bound intermediates.(B)The powdered platinum increases the temperature of the reaction mixture, increasing the rate constant.(C)The powdered platinum changes the overall reaction enthalpy, making the reaction more thermodynamically favorable.(D)The powdered platinum is consumed more rapidly than foil platinum, which drives the reaction forward faster.

10. [Skill: 5.B | Topic: 5.2] A student investigates the reaction $2NO(g)+O_2(g)\rightarrow 2NO_2(g)$. The student monitors the reaction by measuring the absorbance of $NO_2$ versus time using a colorimeter and determines the initial rate from the initial slope of a plot of $[NO_2]$ versus time. The student performs several trials at the same temperature and obtains the following initial-rate data: Trial 1: $[NO]=0.10\ \text{M}$, $[O_2]=0.10\ \text{M}$, initial rate $=0.20\ \text{M}\,\text{s}^{-1}$ Trial 2: $[NO]=0.20\ \text{M}$, $[O_2]=0.10\ \text{M}$, initial rate $=0.80\ \text{M}\,\text{s}^{-1}$ Trial 3: $[NO]=0.10\ \text{M}$, $[O_2]=0.20\ \text{M}$, initial rate $=0.40\ \text{M}\,\text{s}^{-1}$ Which of the following rate laws is most consistent with the data?

(A)$\text{rate}=k[NO][O_2]$(B)$\text{rate}=k[NO]^2$(C)$\text{rate}=k[NO]^2[O_2]$(D)$\text{rate}=k[NO][O_2]^2$

11. [Skill: 5.A | Topic: 5.2] For a reaction at a particular temperature, the experimentally determined rate law is $\text{rate}=k[A][B]^2$. In a trial, the initial concentrations are $[A]=0.150\ \text{M}$ and $[B]=0.200\ \text{M}$, and the initial rate is $3.60\times10^{-3}\ \text{M}\,\text{s}^{-1}$. Which of the following gives the correct value of $k$ and its units?

(A)$k=0.600\ \text{s}^{-1}$(B)$k=0.600\ \text{M}^{-2}\,\text{s}^{-1}$(C)$k=6.00\ \text{M}^{-1}\,\text{s}^{-1}$(D)$k=6.00\times10^{-1}\ \text{M}^{2}\,\text{s}^{-1}$

Refer to the figure below.

12. **1. [Skill: 4.A | Topic: 5.3]** A student studies the decomposition of $NOCl(g)$ at constant temperature: $2NOCl(g) \rightarrow 2NO(g) + Cl_2(g)$ The student generates three plots using the same experimental data. Only one plot is linear. [Image Cue]: Multi-panel graph set (three small graphs labeled I, II, III), Title: "Testing Integrated Rate Law Plots for $[NOCl]$", x-axis on all panels: time (s). Panel I y-axis: $[NOCl]$ (M) showing a curved decrease (not linear). Panel II y-axis: $\ln[NOCl]$ showing a straight line with best-fit slope $-4.0\times 10^{-2}\ \text{s}^{-1}$ and y-intercept near $\ln(0.80)$. Panel III y-axis: $1/[NOCl]$ (M$^{-1}$) showing curvature (not linear). Based on the linear plot, which of the following correctly identifies the reaction order with respect to $NOCl$ and the value of the rate constant, $k$?

(A)First order; $k = 4.0\times 10^{-2}\ \text{s}^{-1}$(B)First order; $k = 4.0\times 10^{-2}\ \text{M}^{-1}\text{s}^{-1}$(C)Second order; $k = 4.0\times 10^{-2}\ \text{s}^{-1}$(D)Zeroth order; $k = 4.0\times 10^{-2}\ \text{M}\text{s}^{-1}$

13. **2. [Skill: 2.A | Topic: 5.3]** A radioactive isotope $X$ decays to a stable product according to the equation below. $X \rightarrow Y$ A student measures the mass of $X$ remaining in a sealed container over time. | Time (days) | Mass of $X$ remaining (mg) | |---:|---:| | 0 | 80 | | 10 | 40 | | 20 | 20 | | 30 | 10 | Assuming the decay follows first-order kinetics, what time is required for the mass of $X$ to reach 5 mg?

(A)35 days(B)40 days(C)50 days(D)80 days

14. [Skill: 2.A | Topic: 5.4] A proposed reaction mechanism contains the elementary step $NO_2(g) + CO(g) \rightarrow NO(g) + CO_2(g)$ Which of the following rate law expressions is consistent with this step being elementary?

(A)$\text{rate} = k[NO_2][CO]$(B)$\text{rate} = k\dfrac{[NO_2]}{[CO]}$(C)$\text{rate} = k[NO_2]^2[CO]$(D)$\text{rate} = k[NO][CO_2]$

15. [Skill: 4.A | Topic: 5.4] The overall reaction $2NO(g) + Br_2(g) \rightarrow 2NOBr(g)$ is studied experimentally, and the rate law is determined to be $\text{rate} = k[NO]^2[Br_2]$. A student concludes that the overall reaction occurs in a single elementary step. Which statement best evaluates the student’s conclusion?

(A)The conclusion is justified because the experimental rate law matches the rate law predicted from the stoichiometry of an elementary step.(B)The conclusion is not justified because even though the rate law matches what a termolecular elementary step would predict, simultaneous collisions of three particles are rare and a multistep mechanism could also produce the same rate law.(C)The conclusion is not justified because the stoichiometric coefficients of reactants in an elementary step can never appear as exponents in a rate law.(D)The conclusion is justified because termolecular elementary steps are more common than bimolecular elementary steps.

Refer to the figure below.

16. **1. [Skill: 4.A | Topic: 5.5]** A student models the effect of temperature on the rate of an elementary gas-phase reaction using Maxwell–Boltzmann energy distributions. > **[Image Cue]:** Graph, “Maxwell–Boltzmann Distributions at Two Temperatures,” x-axis labeled “Kinetic Energy,” y-axis labeled “Number of molecules.” Two curves are shown: curve 1 is taller and narrower with a lower most-probable energy (lower temperature, T1), and curve 2 is shorter and broader shifted to the right (higher temperature, T2). A vertical line labeled $E_a$ appears to the right of both peaks. The area to the right of $E_a$ is visibly larger under curve 2 than under curve 1. Which statement best explains why the reaction rate is greater at $T_2$ than at $T_1$?

(A)At $T_2$, a greater fraction of molecules have kinetic energy greater than or equal to $E_a$, so a greater fraction of collisions can be successful.(B)At $T_2$, all molecules have kinetic energy greater than or equal to $E_a$, so every collision results in reaction.(C)At $T_2$, the activation energy $E_a$ decreases because higher temperature lowers the energy barrier for the reaction.(D)At $T_2$, the orientation requirement for reaction is eliminated, so collisions no longer need to be properly aligned.

17. **2. [Skill: 1.B | Topic: 5.5]** Consider the elementary reaction $A + BC \rightarrow AB + C$ occurring in the gas phase. A student proposes: “If we increase the concentration (partial pressure) of $A$, the reaction rate increases because a greater percentage of collisions will have the correct orientation.” Which of the following best evaluates the student’s claim using the collision model?

(A)The claim is correct; increasing the concentration of $A$ increases the fraction of properly oriented collisions.(B)The claim is incorrect; increasing the concentration of $A$ increases the collision frequency, but the fraction of collisions with correct orientation is not increased by concentration alone.(C)The claim is correct; increasing the concentration of $A$ increases the activation energy, which forces the reaction to occur only when orientation is correct.(D)The claim is incorrect; increasing the concentration of $A$ decreases the collision frequency because particles have less space to move.

Refer to the figure below.

18. [Skill: 2.B | Topic: 5.6] A student analyzes an elementary reaction using the reaction energy profile shown. [Image Cue]: Reaction energy profile diagram titled "Energy vs. Reaction Coordinate". y-axis: Potential energy (kJ/mol). x-axis: Reaction coordinate. Reactants plateau at 20 kJ/mol, curve rises to a maximum (transition state) at 75 kJ/mol, then decreases to products plateau at 10 kJ/mol. Based on the diagram, what are the activation energy, $E_a$, for the forward reaction and the overall enthalpy change, $\Delta H$, for the reaction?

(A)$E_a = 55\ \text{kJ/mol}$ and $\Delta H = -10\ \text{kJ/mol}$(B)$E_a = 95\ \text{kJ/mol}$ and $\Delta H = -10\ \text{kJ/mol}$(C)$E_a = 55\ \text{kJ/mol}$ and $\Delta H = +10\ \text{kJ/mol}$(D)$E_a = 65\ \text{kJ/mol}$ and $\Delta H = -55\ \text{kJ/mol}$

Refer to the figure below.

19. [Skill: 4.A | Topic: 5.6] Two runs of the same elementary reaction are performed at different temperatures. A student is provided the following Maxwell–Boltzmann distributions of molecular kinetic energies at $T_1$ and $T_2$, where $T_2 > T_1$. The vertical line on the plot marks the minimum kinetic energy required for a collision to lead to reaction (the threshold associated with reaching the transition state). [Image Cue]: Maxwell–Boltzmann distribution plot titled "Distribution of Molecular Kinetic Energies". x-axis: Kinetic energy. y-axis: Number of molecules. Two curves: a taller, narrower curve labeled $T_1$ peaking at lower energy; a shorter, broader curve labeled $T_2$ shifted to higher energy. A vertical line at a fixed kinetic energy labeled "Threshold energy to reach transition state". The area under the curve to the right of the line is visibly larger for $T_2$ than for $T_1$. Which statement best explains why the reaction rate increases when the temperature is raised from $T_1$ to $T_2$?

(A)At higher temperature, a greater fraction of collisions have enough energy to reach the transition state, increasing the rate constant as described by the Arrhenius equation.(B)At higher temperature, the activation energy of the reaction decreases, so fewer collisions are needed to form products.(C)At higher temperature, the potential energy of the products increases, making the reaction more thermodynamically favorable and therefore faster.(D)At higher temperature, the reaction becomes exothermic, and exothermic reactions always occur faster than endothermic reactions.

20. 1. [Skill: 1.A | Topic: 5.7] A proposed mechanism for the decomposition of hydrogen peroxide is shown. Step 1: $H_2O_2(aq) + I^-(aq) \rightarrow H_2O(l) + IO^-(aq)$ Step 2: $H_2O_2(aq) + IO^-(aq) \rightarrow H_2O(l) + O_2(g) + I^-(aq)$ Which choice correctly identifies the reaction intermediate and the catalyst in the mechanism?

(A)$IO^-$ is the intermediate, and $I^-$ is the catalyst.(B)$I^-$ is the intermediate, and $IO^-$ is the catalyst.(C)$H_2O_2$ is the intermediate, and $IO^-$ is the catalyst.(D)$O_2$ is the intermediate, and $H_2O$ is the catalyst.

21. 2. [Skill: 3.B | Topic: 5.7] The overall reaction is $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$. Two mechanisms are proposed: Mechanism 1 Step 1: $NO(g) + O_2(g) \rightarrow NO_3(g)$ Step 2: $NO_3(g) + NO(g) \rightarrow 2NO_2(g)$ Mechanism 2 Step 1: $NO(g) + NO(g) \rightarrow N_2O_2(g)$ Step 2: $N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g)$ In an experiment, a short-lived species with a mass spectrum consistent with $NO_3(g)$ is detected during the reaction. Which conclusion is best supported by this observation?

(A)The observation supports Mechanism 1 over Mechanism 2 because $NO_3$ is an intermediate in Mechanism 1 but not in Mechanism 2.(B)The observation supports Mechanism 2 over Mechanism 1 because $NO_3$ must be a catalyst that increases the reaction rate.(C)The observation supports both mechanisms equally because intermediates are always present in all mechanisms for the same overall reaction.(D)The observation rules out both mechanisms because detection of $NO_3$ means it must appear in the overall balanced equation.

22. [Skill: 4.A | Topic: 5.8] A reaction has the overall equation $2\,NO_2(g)+F_2(g)\rightarrow 2\,NO_2F(g)$. A proposed mechanism is shown. Step 1 (slow): $NO_2(g)+F_2(g)\rightarrow NO_2F(g)+F(g)$ Step 2 (fast): $F(g)+NO_2(g)\rightarrow NO_2F(g)$ Assuming the first step is rate-limiting and each step is irreversible, which of the following is the predicted rate law for the overall reaction?

(A)$\text{rate}=k[NO_2][F_2]$(B)$\text{rate}=k[NO_2]^2[F_2]$(C)$\text{rate}=k[F]$(D)$\text{rate}=k[NO_2]^2[F_2]^2$

23. [Skill: 2.A | Topic: 5.8] The overall reaction is $H_2O_2(aq)+2I^-(aq)+2H^+(aq)\rightarrow I_2(aq)+2H_2O(l)$. A proposed mechanism is shown. Step 1 (slow): $H_2O_2(aq)+I^-(aq)\rightarrow HOI(aq)+OH^-(aq)$ Step 2 (fast): $HOI(aq)+I^-(aq)+H^+(aq)\rightarrow I_2(aq)+H_2O(l)$ Step 3 (fast): $OH^-(aq)+H^+(aq)\rightarrow H_2O(l)$ Assuming the first step is rate-limiting and each step is irreversible, which of the following is the predicted rate law for the overall reaction?

(A)$\text{rate}=k[H_2O_2][I^-]$(B)$\text{rate}=k[H_2O_2][I^-]^2$(C)$\text{rate}=k[H_2O_2][I^-][H^+]$(D)$\text{rate}=k[HOI][H^+]$

24. **1.** [Skill: 4.A | Topic: 5.9] A student proposes the following mechanism for the overall reaction $A + B + C \rightarrow \text{products}$. **Step 1 (fast, reversible):** $A + B \rightleftharpoons I$ (forward rate constant $k_1$, reverse rate constant $k_{-1}$) **Step 2 (slow):** $I + C \rightarrow \text{products}$ (rate constant $k_2$) Assuming the pre-equilibrium approximation applies to Step 1, which of the following is the correct rate law for the overall reaction in terms of $[A]$, $[B]$, and $[C]$?

(A)$\text{rate} = k_2[I][C]$(B)$\text{rate} = k_2[A][B][C]$(C)$\text{rate} = \dfrac{k_1k_2}{k_{-1}}[A][B][C]$(D)$\text{rate} = \dfrac{k_1}{k_{-1}}[A][B]$

25. **2.** [Skill: 4.A | Topic: 5.9] For the overall reaction $2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)$, the following mechanism is proposed. **Step 1 (fast, reversible):** $NO + Cl_2 \rightleftharpoons NOCl_2$ (forward rate constant $k_1$, reverse rate constant $k_{-1}$) **Step 2 (slow):** $NOCl_2 + NO \rightarrow 2NOCl$ (rate constant $k_2$) Assuming Step 1 establishes a pre-equilibrium, which rate law is predicted by the mechanism?

(A)$\text{rate} = k[NO][Cl_2]$(B)$\text{rate} = k[NO]^2[Cl_2]$(C)$\text{rate} = k[NOCl_2][NO]$(D)$\text{rate} = k[Cl_2]^2$

B. Short Answer Questions (1 Question)

Answer all parts of each question. Answers must be in essay form. Outlines or lists alone are not acceptable.

Question 26:

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